- #1
RickRazor
- 17
- 3
- TL;DR Summary
- Missing conceptual detail in optimization problems
Mentor note: For LaTeX here at this site, don't use single $ characters -- they don't work at all. See our LaTeX tutorial from the link at the lower left corner of the input text pane.
I have a function dependent on 4 variables ##f(r_1,r_2,q_1,q)##. I'm looking to minimize this function in the domain ##0\leq r_1 \leq r_2 \leq 1## with respect to the variables ##r_1, r_2## and ##q_1##.
To find the minima, I first solved ##\frac{\partial f}{\partial r_1}=0## and ##\frac{\partial f}{\partial q_1}=0##, giving ##r_1^*(r_2,q)## and ##q_1^*(r_2,q)##. Now I have the function of the form ##f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q).##
Now I solved ##\frac{\partial f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q)}{\partial r_2}=0## for ##r_2^*(q)##.
So, the final function is of the form ##f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## which is fine. Now I see later that ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##.
So, if I directly look for the function ##f(0,r_2,0,q)## and it's minima wrt ##r_2##, it's giving a different result, i.e. I have
##\min_{r_2} f(0,r_2,0,q) \neq f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## even though ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##. Why is this the case? Are there other simple examples?
The function is
##f(r_1,r_2,q_1,q)=3r_1+r_2+q_1^2/r_1+2(q-q_1)^2/(r_2-r_1)## and
##f(0,r_2,0,q)=r_2+2(q-q_1)^2/r_2##
##r_2^*(q)=\sqrt{\frac{2}{3}}q##,
##r_1(r_2^*(q),q)=0## and ##q_1(r_2^*(q),q)=0##.
I have a function dependent on 4 variables ##f(r_1,r_2,q_1,q)##. I'm looking to minimize this function in the domain ##0\leq r_1 \leq r_2 \leq 1## with respect to the variables ##r_1, r_2## and ##q_1##.
To find the minima, I first solved ##\frac{\partial f}{\partial r_1}=0## and ##\frac{\partial f}{\partial q_1}=0##, giving ##r_1^*(r_2,q)## and ##q_1^*(r_2,q)##. Now I have the function of the form ##f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q).##
Now I solved ##\frac{\partial f(r_1^*(r_2,q),r_2,q_1^*(r_2,q),q)}{\partial r_2}=0## for ##r_2^*(q)##.
So, the final function is of the form ##f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## which is fine. Now I see later that ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##.
So, if I directly look for the function ##f(0,r_2,0,q)## and it's minima wrt ##r_2##, it's giving a different result, i.e. I have
##\min_{r_2} f(0,r_2,0,q) \neq f(r_1^*(r_2^*(q),q),r_2^*(q),q_1^*(r_2^*(q),q),q)## even though ##r_1^*(r_2^*(q),q)=0## and ##q_1^*(r_2^*(q),q)=0##. Why is this the case? Are there other simple examples?
The function is
##f(r_1,r_2,q_1,q)=3r_1+r_2+q_1^2/r_1+2(q-q_1)^2/(r_2-r_1)## and
##f(0,r_2,0,q)=r_2+2(q-q_1)^2/r_2##
##r_2^*(q)=\sqrt{\frac{2}{3}}q##,
##r_1(r_2^*(q),q)=0## and ##q_1(r_2^*(q),q)=0##.
Last edited by a moderator: